Đáp án:
A = $\frac{-1}{3}$ + $\frac{1}{3²}$ - $\frac{1}{3³}$ +....+ $\frac{1}{3^{100}}$
=> 3A = 3($\frac{-1}{3}$ + $\frac{1}{3²}$ - $\frac{1}{3³}$ +....+ $\frac{1}{3^{100}}$ )
= -1 + $\frac{1}{3}$ - $\frac{1}{3²}$ + $\frac{1}{3³}$ +....+ $\frac{1}{3^{99}}$
=> 3A + A = 4A = (-1 + $\frac{1}{3}$ - $\frac{1}{3²}$ + $\frac{1}{3³}$ +....+ $\frac{1}{3^{99}}$ ) + ($\frac{-1}{3}$ + $\frac{1}{3²}$ - $\frac{1}{3³}$ +....+ $\frac{1}{3^{100}}$)
= -1 + $\frac{1}{3^{100}}$
=> A = (-1 + $\frac{1}{3^{100}}$):4
Ta có: B = 4|A| + $\frac{1}{3^{100}}$
= 4.|(-1 + $\frac{1}{3^{100}}$):4 | + $\frac{1}{3^{100}}$
= 4.(1 - $\frac{1}{3^{100}}$):4 + $\frac{1}{3^{100}}$
= (1 - $\frac{1}{3^{100}}$) + $\frac{1}{3^{100}}$
= 1
Vậy B = 1
